Sun Sensor Previous

1. Sun SensorPrevious Jairo Alberto Hurtado

2. What is a sun sensor? • The sun sensor tells where the sun is, to help orient the space craft (taking aim to earth). • There are two options to choose the type of sensor: PSD (Position-Sensitive Detector) or CMOS

3. Sensor type

4. Geometric Description 1/3 • Sun light goes through a small hole and impacts on the sensor. • With the position of the spot beam on the sensor we can calculate elevation and azimuth angles.

5. Geometric Description 2/3 • The shape and geometry of the system with the PSD sensor in shown here: The maximum angle due to the geometry for any sensor is the minimum between and .

6. Geometric Description 3/3 • With 66,8° the maximum angle (diagonal), the distance from the pin hole to the PSD sensor is given by: • With h=2,73 mm, the maximum angle of incidence to reach the border of the sensor is given by:

7. Hole size 1/2 • Geometry gives these maximum angles, 58,8° and 66,8° and a distance of h ≈ 2,73 mm. • Now, optical analysis to get the size hole starts with a method of calculating the optimal pinhole diameter. Using the Lord Rayleigh equation, we have: • Where d is diameter, h is focus length (distance from hole to sensor) and λ is the wavelength of light, so:

8. Hole size 2/2 • The minimum “maximum angle” given as parameter is 50°, in that way, we get h ≈ 3,78 mm and the maximum angle (diagonal) will be of 59,3°. • Applying newly the Lord Rayleigh equation: • As the hole’s diameters available are 50 μm ± 5 μm, 100μm ± 5 μm and 200 μm ± 5 μm.(Edmund Optics) • Finally the pin hole diameter choosed is d =100 μm.

9. Final values for h 1/1 • Defined d, we have an interval to selected h given by: - Maximum angle of incidence, 58,8°. (h ≈ 2,7 mm) - Minimum “maximum angle” of incidence, 50°. (h ≈ 3,8 mm). - Because the Lord Rayleigh equation. (h ≈ 2,9 mm). • The final results with a pin hole of 100 μm, using a sensor of 9 mm of length, are:

10. Pin hole geometry 1/2 • Thickness of the pin hole is 25,4 μm. So, the diameter of the beam (nd = new diameter) depends of the angle of incidence:

11. Pin hole geometry 2/2

12. lp2 α/2 lp0 lp1 θ lf2 α/2 α lf0 θ lf1 h Optical geometry 1/5 • Sun has an angular diameter (α=0,5°). • Projection on the sensor when (β=0°) is: • Generalizing, as a β function, the projection on the sensor is given by:

13. Optical geometry 2/5 • Changes on the size of the image of the spot light on the sensor, in function on the angle of incidence β. f(β) = 2h·tan(0,25°) ·[1 + tanβ]

14. lf2 α lf0 lf1 h d lf2’ α lf0’ lf1’ Optical geometry 3/5 • Using d = 100 μm as the pin hole diameter. • Image on the sensor is given for the distance between from lf2 to lf1’.ims = (|lf2 - lf1’|). Ims

15. Optical geometry 4/5 • Figureincluding the variations due to the thickness of the hole and the spread of the beam spot on the sensor, in function on the angle of incidence β.

16. Optical geometry 5/5 • Zoom on the last function with maximum value of β=50°.

17. Diffraction 1/1 • Diffraction pattern on the sensor to 3 mm of distance, with a hole diameter of 100 μm and λ = 960 nm.

18. Error and Resolution 1/3 • Based in data sheet of PSD: - Position Resolution: Typ. 1,5 μm - Position detection error: Typ.150 μm, @ (λ=900 nm, spot light size = 200 μm) Position detection error example of one-dimensional PSD (S4583-04)

19. Error and Resolution 2/3 • Taking the position error as a linear function with maximum value 150 μm. • Where L = distance of the spot from the center, res = distance error. (h token as 3 mm).

20. Error and Resolution 3/3 • Angular error function, taking the position error as a linear function with maximum value 150 μm. • Angular resolution due to position resolution (1,5 μm) is better than 0,03°

22. With a Kodak sensor 1/1 • Applying the same calculations using a Kodak sensor with active area 4,86 mm x 3,66 mm, the results are: • Defined d = 100 μm, we have an interval to selected h given by: - Maximum angle of incidence, 54,6°. (h ≈ 1,3 mm) - Minimum “maximum angle” of incidence, 50°. (h ≈ 1,5 mm). - Because the Lord Rayleigh equation. (h ≈ 3,7 mm). - *Using d = 50 μm, with Lord Rayleigh eq. (h ≈ 0,9 mm). *Light spot narrower.

23. Optical geometry 1/3 • Changes on the size of the image of the spot light on the sensor, in function on the angle of incidence β. f(β) = 2h·tan(0,25°) ·[1 + tanβ]

24. Optical geometry 2/3 • Figureincluding the variations due to the thickness of the hole and the spread of the beam spot on the sensor, in function on the angle of incidence β.

25. Optical geometry 3/3 • Zoom on the last function with maximum value of β=50°.

26. Error and Resolution 1/1 • Pixel size: 7,5 μm. • Resolution: ± 1 pixel With this values, angular resolution due to position resolution (7,5 μm) is always better than 0,3°

28. With a Aptina sensor 1/1 • Applying the same calculations using a Kodak sensor with active area 4,51 mm x 2,88 mm, the results are: • Defined d = 100 μm, we have an interval to selected h given by: - Maximum angle of incidence, 51,4°. (h ≈ 1,15 mm) - Minimum “maximum angle” of incidence, 50°. (h ≈ 1,2 mm). - Because the Lord Rayleigh equation. (h ≈ 3,7 mm). - *Using d = 50 μm, with Lord Rayleigh eq. (h ≈ 0,9 mm). *Light spot narrower.

29. Optical geometry 1/1 • Figuresincluding the variations due to the thickness of the hole and the spread of the beam spot on the sensor, in function on the angle of incidence β.

30. Error and Resolution 1/1 • Pixel size: 6 μm. • Resolution: ± 1 pixel With this values, angular resolution due to position resolution (6 μm) is always better than 0,3°